and 3 others joined a min ago.
0
4.8kviews
A 3 Phase, 10 KVA load has power factor of 0.342. The power is measured by two wattmeter method. Find reading of each wattmeter when i) Power Factor is leading ii) Power Factor is logging
1 Answer
0
1.0kviews
written 7.8 years ago by
teamques10
★ 65k
|
Given
$10\times10^3 VA \\ Pf=0.342 \\ P_{apperent}=10kvk \\ \sqrt3V_LI_L=10^4 \\ V_LI_L=\dfrac{10^4}{\sqrt3}=5773.5VA \\ pf=0.342 \emptyset=\cos_{-1}(0.342)$
1. Pf is lending, then
$\hspace{0.5cm}$ $w_1=V_LI_L \cos(30-\emptyset)=5773 \cos(30-70) \\=4422.76watt \\=4.42kw \\ w_2=V_LI_L \cos(30+\emptyset)=-1002.55watt \\ =-1.0025kw$
2. pf is logging, then
$\hspace{0.5cm}$ $w_1=V_LI_L \cos(30+\emptyset)=5773\cos(30+70) \\ =-1.0025kw \\w_2=V_LI_L\cos(30-\emptyset)=4.42kw$
and 4 others joined a min ago.
ADD COMMENT
EDIT
Please log in to add an answer.